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This question is inspired from the definition of Hausdorff measure. Let $C$ be a closed unit hypercube in $\mathbb R^d$ (side length equal to one, including boundary. The cube itself is at top dimension.). Consider all possible not-necessarily-open countable covers of it.

I would like to find a lower bound estimate for the sum $\sum \text{diam}(U)^d$ over all possible countable covers of $C$. Namely, what is a lower bound estimate for

$$\inf \{\sum_{U\in \mathscr U} \text{diam}(U)^d: \mathscr U~ \text{is a not-necessarily-open countable cover of} ~C \}?$$

The difference here from Hausdorff measure is that I dropped the restriction on the diameter. Note with that restriction, the infimum only gets larger.

I only know this number should be a positive constant that only depends on $d$. But I don't know how to give a naive lower estimate for it. If the precise quantity turns out to be one of the open problems, I am happy to know the latest progress in this.

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    $\begingroup$ Let $\mathscr{U}=\{U_n;n\in\mathbb{N}\}$, and let $d_n$ be the diameter of $U_n$. Then $U_n$ is contained in a closed cube $C_n$ of side $d_n$, and the sequence $C_n$ covers $C$ so $1\leq\sum_n vol(C_n)^d=\sum_{U\in \mathscr U} \text{diam}(U)^d$. $\endgroup$
    – Saúl RM
    Commented Feb 16, 2022 at 23:29
  • $\begingroup$ @SaúlRodríguezMartín Why $1≤∑_n vol(C_n)^d$? I suppose $∑_n vol(C_n)^d$ may be much smaller than $∑_n vol(C_n)$ which is bounded below by $1$ $\endgroup$
    – No One
    Commented Feb 16, 2022 at 23:35
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    $\begingroup$ Oh sorry, I meant $1\leq\sum_n vol(C_n)=\sum_{U\in\mathscr{U}}diam(U)^d$, the first $d$ is a typo $\endgroup$
    – Saúl RM
    Commented Feb 16, 2022 at 23:37

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